Question

in an equilateral triangle ABC D is a point on the side BC such that BD is equal to 1 by 3 BC prove that 9AD square is equal to 7AD square

Answer 1

bd= 1/3bc
draw ae perpendicular on BC .join AD.
in aeb and arc 
aeb is congruent to aec by rhs
so be =ec by cpct
in AED ad square =ae square+DE square
AD square= ab square -be square +de square 
ad square =7bc square /9
9 ad square =7ab square

Answer 2

Here is your solution

Given:-

ABC is an equilateral triangle.

D is point on BC .

so

BD =BC.

To prove:-

9 AD² = 7 AB²

Construction: Draw AE ⊥ BC.

Proof :-


In a ΔABC and ΔACE

AB = AC ( given)

AE = AE (common)

∠AEB = ∠AEC = (Right angle)

∴ ΔABC ≅ ΔACE (By RHS )

∴ ΔABC ≅ ΔACE 

Again,

BE = EC (By C.P.C.T)

BE = EC = BC²

In a right angled ΔADE

AD²= AE2 + DE² ---(1)

In a right angled ΔABE

AB² = AE² + BE² ---(2)

From equation (1) and (2) ;

=) AD²  - AB² =  DE² - BE².

=) AD²  - AB² = (BE – BD)² - BE².

= ) AD²  - AB² = (BC / 2 – BC/3)² – (BC/2)²

= AD2  - AB2 = ((3BC – 2BC/6)² – (BC/2)² 

= AD²  - AB² = (BC² / 36 – BC2 / 4 )

( In a equilateral triangle, All sides are equal to each other)

AB = BC = AC

= ) AD²= AB² + AB²/ 36 – AB² / 4

= )AD² = (36AB² + AB²– 9AB²) / 36

= ) AD² = (28AB²) / 36

=) AD² = (7AB²) / 9

Cross Multiplication here,

= ) 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎

‎


‎Hence,

9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎proved

Hope it helps you